Answer:
is closest to the total surface area of the box
Step-by-step explanation:
we know that
The surface area of the box is equal to

where
B is the area of the base
P is the perimeter of the base
H is the height of the box
we have



<em>Find the area of the base B</em>


<em>Find the perimeter of the base P</em>



<em>Find the surface area</em>



Between 10 and 11, all the others would be to much or to little
The question is missing the figure which is attached below.
Answer:
The last box that has dimensions 8 in × 8 in × 10 in
Step-by-step explanation:
Given:
Volume of the soil = 924 cubic inches.
There are four different types of boxes that need to be checked whether they can accommodate all of the soil or not.
The volume of the box must be at least 924 cubic inches to accommodate all of the soil.
Now, volume of the first box is given as:

Volume of the second box is given as:

Volume of the third box is given as:

Volume of the fourth box is given as:

Therefore, only the volume of the fourth box is less than the total volume of the soil. So, last box is the correct option.
Answer:
105.42
Step-by-step explanation:
259.63
<u>154.21</u>
105.42
9°, complementary angles always add up to 90°